Answer:
19559303watts
Explanation:
The field inside a solenoid is
B =µ0*N*I/h
where N is the number of turns, I the current in each turn, and h the length.
If the wire's diameter is d, then the number of turns is N = h/d so the equation becomes
B = µ0*I/d
I = B*d/µ0
B = 7.8*10^-3 T, d = 0.001 m, µ0 = 4*π*10^-7 H/m
I = 8*10^-3*0.001/(4*π*10^-17))
I = 6.366A
For power
We need to know ρ, the resistivity of the wire. The resistance of the coil R = ρ*L/A where L is the total wire length and A the area.
L = N*π*D, were D = diameter of the solenoid and N the no of turns. As above, N = h/d so L = h*π*D/d. The wire area is π*d²/4 so the resistance is
R = ρ*h*π*D/d*(π*d²/4) = 4*ρ*h*D/d³
The power is I²*R = (B*d/µ0)²*4*ρ*h*D/d³ = (B/µ0)²*4*ρ*h*D/d
For copper ρ = 16.8*10^-9 Ω-m and
D = 0.10 m, h = 0.722 m
P = (6.366 / 4 * π * 10^-7)^2 * 4 * 16.8 * 10 ^-9 * 0.722 * 0.1/0.001
P = 19559303watts
P = 20MW
Answer:
The power is [tex]174W[/tex]
Explanation:
diameter: [tex]d=0.1cm=0.1*10^{-2} m[/tex]
Radius: [tex]r=\frac{d}{2} ,r=5*10^{-4} m[/tex]
Length: [tex]l=63.3cm=0.633m[/tex]
The n will be:
[tex]n=\frac{1m}{0.1*10^{-2} m} \\n=1000turns/m[/tex]
N=n*l
N=1000*0.633
[tex]N=6.33*10^{2} turns[/tex]
To find current:
[tex]I=\frac{B}{n_{0}*n } \\I=\frac{8*10^{3} }{4\pi*1000 } \\I=6.37A[/tex]
To find L:
[tex]L=N*\pi d\\L=(6.33*10^{2} )*(\pi .10*10^{-2}m) \\L=63.3\pi[/tex]
Now we can apply resistivity formula to find resistance:
[tex]R=p\frac{L}{A} \\R=\frac{(1.7*10^{-8}) *(63.3\pi )}{\pi(5*10^{-4} \\)^{2} } \\[/tex]
[tex]R=4.3[/tex]Ω
Now to find the power:
[tex]Power=I^{2} R\\Power=6.37^{2}*4.3\\ Power=174W[/tex]
